HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Use the equation Sn=n2(2a+(n−1)d)
to write an equation using the information about the sum of the first
22 terms. Then write out another equation based on the difference
between the sixth and third terms.
STEP: Use the formula Sn=n2(2a+(n−1)d) to write an equation for the sum of the first 22 terms
[−1 point ⇒ 5 / 6 points left]
There is a lot of information in the question. We
will start with the statement that the sum of the first 22 terms is
1210. This information fits with the formula Sn=n2(2a+(n−1)d), which connects the sum of n terms of an arithmetic series to the values of a and d.
Substitute in what we know to see what the equation can give us:
Sn=n2(2a+(n−1)d)1210=222(2a+(22−1)d)2420=22(2a+21d)110=2a+21d
For the moment we are stuck because there are two variables in the equation (a and d), so we need to let this equation sit for the moment while we see what else we can get out of the question.
STEP: Use the information about the other terms to write another equation
[−2 points ⇒ 3 / 6 points left]
In the work above, we got an equation with two
variables, so the plan for the question is now clear: find another
equation with the variables a and d
so that we can solve the equations simultaneously. Let us try to do
that using the information in the question about the other terms.
In this case, the question states that the "difference between the third and sixth terms ... is 18". We can write this as an equation: T6−T3=18. Using the relationship Tn=a+(n−1)d, we can write this equation in terms of a and d:
T6−T3=18(a+5d)−(a+2d)=183d=18d=6
This is fantastic: the a's cancelled so we could actually find the value of the common difference, d.
STEP: Use the first equation and the value of d to find a
[−1 point ⇒ 2 / 6 points left]
We now have the two equations:
Substitute d=6 into equation (1) to find the value of a.
2a+21d=1102a+21(6)=1102a=−16a=−8
We now have the value of the first term, T1=a=−8, and the common difference, d=6.
STEP: Calculate the values of the second and third terms
[−2 points ⇒ 0 / 6 points left]
Now use the values of a=−8 and d=6 to calculate the values of T2 and T3.
T2=a+d=−8+(6)=−2T3=a+2d=−8+2(6)=4
The first three terms of the sequence are: T1=−8,T2=−2,T3=4.
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